A die is thrown twice. Let A be the event of | vedclass

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(A) When a die is thrown twice,the total number of outcomes is $6 	imes 6 = 36$. Event $A$ is getting a prime number on the first throw. The prime nu

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) When a die is thrown twice,the total number of outcomes is $6 	imes 6 = 36$. Event $A$ is getting a prime number on the first throw. The prime numbers on a die are ${2, 3, 5}$. So,$P(A) = \frac{3}{6} = \frac{1}{2}$. Event $B$ is getting an even number on the second throw. The even numbers on a die are ${2, 4, 6}$. So,$P(B) = \frac{3}{6} = \frac{1}{2}$. Event $\overline{B}$ is the complement of $B$,which means getting an odd number on the second throw. The odd numbers are ${1, 3, 5}$. So,$P(\overline{B}) = 1 - P(B) = 1 - \frac{1}{2} = \frac{1}{2}$. Since the two throws are independent events,the occurrence of $A$ does not depend on the occurrence of $\overline{B}$. Therefore,$P(A / \overline{B}) = P(A) = \frac{1}{2}$.

$A$ die is thrown twice. Let $A$ be the event of getting a prime number when the die is thrown first time and $B$ be the event of getting an even number when the die is thrown second time. Then $P(A / \overline{B})=$

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